A particle starts at time t=0 and moves along the xaxis so that its position at any time t≥0 is given by
x(t)=(t1)^3*(2t3).
a) For what values of t is the velocity of the particle less than zero?
v(t) = x’(t) = 3(t1)^2*(2t3)+2(t1)^3
= (t1)^2*(3*(2t3)+2(t1))
= (t1)^2*(6t9+2t2)
= (t1)^2*(8t11)
v(t) has odd degree (3), so end behaviors are opposite.
t → ∞ ==> v(t) → ∞
end behavior is up to the right and down to the left
zeros are 1, with multiplicity 2, and 11/8=1+3/8
going from right to left, v in Quadrant I comes down and crosses xaxis at 11/8, goes down to a turning point and back up to touch and bounce off xaxis at 1, then goes in negative direction forever.
–––––0––––––0++++> v(t)
––––––––––––––> t
0 1 11/8
So v(t) < 0 for x < 1 and 1 < x < 11/8.
In interval notation: (–∞,1)U(1,11/8)
b) Find the value of t when the particle is moving and the acceleration is zero.
v(t) = (t1)^2*(8t11).
a(t) = v’(t) = 2(t1)(8t11)+8(t1)^2
= 2(t1)((8t11)+4(t1))
= 2(t1)(12t15)
= 6(t1)(4t5)
a(t) is a parabola opening up
with zeros at t=1 and t=5/4=1+2/8.
+++++0––––––0++++++++++++> a(t)
–––––0––––––––––––––0++++> v(t)
–––––––––––––––––––––> t
0 1 (1+2/8) (1+3/8)
At t = 5/4 = 1+2/8 the particle is moving, v(1+2/8) ≠ 0, and the acceleration is zero, a(1+2/8) = 0.
4/7/2014

Steve S.